\(5.03 \times 10^{-3}= C \times 0.083 \times 300\)
\(C =0.202 \times 10^{-3}\, M\)
Moles of protein \(=0.202 \times 10^{-3} \times 0.5\)
\(=10^{-4} \times 1.01\)
\(1.01 \times 10^{-4}=\frac{2.5}{ M }\)
\(M (\) molar mass of protein \()=24752\)
\(\therefore\) No. of glycine units \(=\frac{24752}{75}=330.03\)
[આપેલ $K_b (H_2O) = 0.52\, K\, kg\, mol^{-1}]$